Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int L = 17;
int lv, h[N];
int dp[L][N];
vector<int> mn[L][N];
vector<int> a[N], g[N];
vector<int> merge(vector<int>& x, vector<int>& y, int s) {
  vector<int> z;
  z.reserve(s);
  int i = 0, j = 0;
  while (z.size() < s and i < x.size() and j < y.size()) {
    if (x[i] > y[j])
      z.emplace_back(y[j++]);
    else
      z.emplace_back(x[i++]);
  }
  while (z.size() < s and i < x.size()) z.emplace_back(x[i++]);
  while (z.size() < s and j < y.size()) z.emplace_back(y[j++]);
  return z;
}
void dfs(int u, int p) {
  dp[0][u] = p;
  mn[0][u] = a[u];
  for (int i = 1; i < lv; ++i) {
    mn[i][u] = merge(mn[i - 1][u], mn[i - 1][dp[i - 1][u]], 10);
    dp[i][u] = dp[i - 1][dp[i - 1][u]];
  }
  for (int v : g[u])
    if (v != p) {
      h[v] = h[u] + 1;
      dfs(v, u);
    }
}
vector<int> lca(int u, int v, int s) {
  vector<int> re;
  if (h[u] < h[v]) swap(u, v);
  for (int i = lv - 1; i >= 0; --i)
    if (h[u] - h[v] >> i & 1) {
      re = merge(re, mn[i][u], s);
      u = dp[i][u];
    }
  if (u == v) {
    re = merge(re, mn[0][u], s);
    return re;
  }
  for (int i = lv - 1; i >= 0; --i)
    if (dp[i][u] != dp[i][v]) {
      re = merge(re, mn[i][u], s);
      re = merge(re, mn[i][v], s);
      u = dp[i][u];
      v = dp[i][v];
    }
  if (u != v) {
    re = merge(re, mn[0][u], s);
    re = merge(re, mn[0][v], s);
  }
  re = merge(re, mn[0][dp[0][u]], s);
  return re;
}
int main() {
  int n, m, q, u, v, s;
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i < n; ++i) {
    scanf("%d %d", &u, &v);
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &u);
    a[u].emplace_back(i);
  }
  for (lv = 0; (1 << lv) <= n; ++lv)
    ;
  dfs(1, 0);
  while (q--) {
    scanf("%d %d %d", &u, &v, &s);
    vector<int> re = lca(u, v, s);
    printf("%lu", re.size());
    for (int x : re) printf(" %d", x);
    printf("\n");
  }
  return 0;
}